Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

int n[100000], m[100000];
vector<int>rinsetu[100000];
signed main() {
	unordered_map<int, int>X, Y;
	int a, b, c, d; scanf("%d%d%d%d", &a, &b, &c, &d); c--; d--;
	for (int e = 0; e < b; e++) {
		int f, g; scanf("%d%d", &f, &g);
		f--; g--;
		rinsetu[f].push_back(g);
		rinsetu[g].push_back(f);
	}
	memset(n, -1, sizeof(n));
	memset(m, -1, sizeof(m));
	n[c] = 0; m[d] = 0;
	queue<int>Q;
	Q.push(c);
	while (Q.size()) {
		int e = Q.front(); Q.pop();
		X[n[e]]++;
		for (int f : rinsetu[e]) {
			if (n[f] == -1) {
				n[f] = n[e] + 1;
				Q.push(f);
			}
		}
	}
	Q.push(d);
	while (Q.size()) {
		int e = Q.front(); Q.pop();
		Y[m[e]]++;
		for (int f : rinsetu[e]) {
			if (m[f] == -1) {
				m[f] = m[e] + 1;
				Q.push(f);
			}
		}
	}
	long long ans = 0;
	for (int i = 0; i < a; i++) {
		if(m[i]!=-1)ans += X[n[d] - m[i] - 2];
		if(n[i]!=-1)ans += Y[m[c] - n[i] - 2];
	}
	printf("%lld\n", ans / 2);
}